math question

[busmaster11]

<< P(x>=1) = 1 - P(x=0).

The binomial functional form is P(x) = (n!/(r!(n-r)!)) * p^r * (1-p)^(n-r)
where n is the number of events (100, since you are drawing 100 times), r is the number of possible successful numbers (in this case, zero, as we are testing for the probability of getting #50 zero times), and p is the probability of success on any one trial (1% or .01, since there is a 1/100 chance of success).

Plug and chug... (100!/(0!(100-0)!)) * (0.01)^0 * (1-0.01)^(100-0)
(100! / 100!) * 1 * (0.99)^100 = .366 = 36.6%.

In other words, the chances of getting ball #50 zero times in 100 draws is equal to 36.6%. Then, 1 minus that result tells you the chances of getting ball #50 AT LEAST once. So, 1 - .366 = .634 = 63.4%.

Got it? >>



Excellent. I'm impressed.
I suspect any of us who have taken college stats ought to know this though few of us know/remember...

 

[tikwanleap]

the first question asks for the probablity of the number 50 appearing exactly once in a sequence of 100 trials.

In one trial:
Probability of getting a 50 = 1/100
Probability of getting anything else = 99/100

We need the number 50 appearing once so that's 1/100.
Then we need any number except 50 appearing the other 99 times, so that's (99/100)^99

So the answer would be = (1/100)^1 * (99/100)^99 = ~0.00370 or 0.370%

I think that's right, holler if I'm wrong.

 

[tikwanleap]

For the at least once problem, i would do it like the others have already explained.

Find the probability that the number 50 never appears in a sequence of 100 trials => (99/100)^100 = ~0.366

then subtract that from 1 => 1 - 0.366 = 0.634 = 63.4%

 

[tikwanleap]

Twice and three times follows the same method.

Twice:
(1/100)^2 * (99/100)^98 = ~ 0.0000373 or 0.00373%

Three times:
(1/100)^3 * (99/100)^97 = ~ 0.000000377 or 0.0000377%

 

[GermyBoy]

EDIT: It seems like you already have the answer.

 

[b0mbrman]

<< OK. One hundred pingpong balls in a lottery drum numbered one to a hundred. In a hundred turns, what is the probability you'll get the number "50" (just to pick an arbitrary number) exactly once?
How about atleast once?
How about twice? How about three times?
How about not at all?
And how did you calculate it? >>


I'm sure these have all been answered already but I would have liked to have tried them out...anyhow, good luck to all involved

 

[busmaster11]

<< Twice and three times follows the same method.

Twice:
(1/100)^2 * (99/100)^98 = ~ 0.0000373 or 0.00373%

Three times:
(1/100)^3 * (99/100)^97 = ~ 0.000000377 or 0.0000377% >>



Wait, you're saying once would be, .37%, twise is .00373%, 3x is .0000377%... And it diminishes further after that. Correct me if I'm wrong, but shouldn't 1x be the most likely result, and every other possibility diminishes after that? If so, you'll never even hit 1%, much less the 64.3% you agree with.

 

[busmaster11]

<<

<< OK. One hundred pingpong balls in a lottery drum numbered one to a hundred. In a hundred turns, what is the probability you'll get the number "50" (just to pick an arbitrary number) exactly once?
How about atleast once?
How about twice? How about three times?
How about not at all?
And how did you calculate it? >>


I'm sure these have all been answered already but I would have liked to have tried them out...anyhow, good luck to all involved >>



Not quite yet, so you still have your shot. But Arschlock pretty much gave us the biggest one.